Question 1

Find the margin of error for the 95% confidence interval used to estimate the population proportion
-n = 169, x = 107

Accepted Answer

A) 0.127 
B) 0.0654 
C) 0.00269 
D) 0.0727 
A) 0.127 
B) 0.0654 
C) 0.00269 
D) 0.0727

Question 2

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
-Weights of eggs: $95 \%$ confidence; $\mathrm { n } = 22 , \overline { \mathrm { x } } = 1.77 \mathrm { oz } , \mathrm { s } = 0.47 \mathrm { oz }$

Accepted Answer

A)  $0.36 \mathrm { oz } < \sigma < 0.65 \mathrm { oZ }$ 
B)  $0.37 \mathrm { oz } < \sigma < 0.61 \mathrm { oz }$ 
C)  $0.36 \mathrm { oz } < \sigma < 0.67 \mathrm { oz }$ 
D)  $0.38 \mathrm { oz } < \sigma < 0.63 \mathrm { oz }$ 
A)  $0.36 \mathrm { oz } < \sigma < 0.65 \mathrm { oZ }$ 
B)  $0.37 \mathrm { oz } < \sigma < 0.61 \mathrm { oz }$ 
C)  $0.36 \mathrm { oz } < \sigma < 0.67 \mathrm { oz }$ 
D)  $0.38 \mathrm { oz } < \sigma < 0.63 \mathrm { oz }$

Question 3

Use the margin of error, confidence level, and standard deviation ? to find the minimum sample size required to estimate an unknown population mean µ
-A group of 51 randomly selected students have a mean score of $25.4$ with a standard deviation of $3.1$ on a placement test. What is the 90 percent confidence interval for the mean score, $\mu$, of all students taking the test?

Accepted Answer

A)  $24.3 < \mu < 26.5$ 
B)  $24.4 < \mu < 26.4$ 
C)  $24.7 < \mu < 26.1$ 
D)  $24.5 < \mu < 26.3$ 
A)  $24.3 < \mu < 26.5$ 
B)  $24.4 < \mu < 26.4$ 
C)  $24.7 < \mu < 26.1$ 
D)  $24.5 < \mu < 26.3$

Question 4

Provide an appropriate response.
-$$\text { Why is } s ^ { 2 } \text { the best point estimate of } \sigma ^ { 2 } \text { ? }$$

Accepted Answer

The answer of Provide an appropriate response.
-\[\text { Why is...

Question 5

Find the margin of error
-$90 \% ; n = 9 ; \sigma = 4.2 ;$ population appears to be very skewed.

Accepted Answer

A)  $\mathrm { z } _ { \alpha / 2 } = 2.896$ 
B)  $\mathrm { z } _ { \alpha / 2 } = 2.306$ 
C)  $\mathrm { z } _ { \alpha / 2 } = 2.365$ 
D)  Neither the normal nor the $t$ distribution applies. 
A)  $\mathrm { z } _ { \alpha / 2 } = 2.896$ 
B)  $\mathrm { z } _ { \alpha / 2 } = 2.306$ 
C)  $\mathrm { z } _ { \alpha / 2 } = 2.365$ 
D)  Neither the normal nor the $t$ distribution applies.

Question 6

The Bide-a-While efficiency hotel, which caters to business workers who stay for extended periods of time (weeks or months), offers room service. In a small study of 35 randomly selected room service orders, the 95% confidence interval for mean delivery time for room service is 24.8 < µ < 29.6 minutes. The marketing director is trying to determine if she can advertise "room service in under 30 minutes, or the order is free." How would you advise her?

Accepted Answer

We are 95% sure that the interval 24.8 <

Question 7

Solve the problem.
-Of 243 employees selected randomly from one company, 14.81% of them commute by carpooling. Construct a 90% confidence interval for the true percentage of all employees of the company who carpool.

Accepted Answer

A) 9.50% < p < 20.1% 
B) 10.3% < p < 19.3% 
C) 8.94% < p < 20.7% 
D) 11.1% < p < 18.6% 
A) 9.50% < p < 20.1% 
B) 10.3% < p < 19.3% 
C) 8.94% < p < 20.7% 
D) 11.1% < p < 18.6%

Question 8

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
-To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s = 0.16. Find the 95% confidence interval for the population standard deviation ?.

Accepted Answer

A)  $0.13 < \sigma < 0.22$ 
B)  $0.15 < \sigma < 0.21$ 
C)  $0.12 < \sigma < 0.24$ 
D)  $0.11 < \sigma < 0.25$ 
A)  $0.13 < \sigma < 0.22$ 
B)  $0.15 < \sigma < 0.21$ 
C)  $0.12 < \sigma < 0.24$ 
D)  $0.11 < \sigma < 0.25$

Question 9

The following confidence interval is obtained for a population proportion, p: 0.855 < p < 0.897 Use these confidence interval limits to find the margin of error, E.

Accepted Answer

A) 0.021 
B) 0.876 
C) 0.022 
D) 0.042 
A) 0.021 
B) 0.876 
C) 0.022 
D) 0.042

Question 10

Solve the problem.
-A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 6 percentage points in either direction from what would have been obtained by interviewing all voters in the United States." Find the sample size suggested by this statement.

Accepted Answer

A) 268 
B) 19 
C) 461 
D) 378 
A) 268 
B) 19 
C) 461 
D) 378

Question 11

Use the confidence level and sample data to find the margin of error E
-Replacement times for washing machines: $90 \%$ confidence; $\mathrm { n } = 36 , \bar{\mathrm { x} } = 10.0$ years, $\sigma = 2.1$ years

Accepted Answer

A)  $0.4$ years 
B)  $6.0$ years 
C)  $0.1$ years 
D)  $0.6$ years 
A)  $0.4$ years 
B)  $6.0$ years 
C)  $0.1$ years 
D)  $0.6$ years

Question 12

Solve the problem.
-Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all such items that are defective.

Accepted Answer

A) 0.0154 < p < 0.0540 
B) 0.0118 < p < 0.0576 
C) 0.0345 < p < 0.0349 
D) 0.0110 < p < 0.0584 
A) 0.0154 < p < 0.0540 
B) 0.0118 < p < 0.0576 
C) 0.0345 < p < 0.0349 
D) 0.0110 < p < 0.0584

Question 13

Solve the problem.
-When 306 college students are randomly selected and surveyed, it is found that 115 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.

Accepted Answer

A) 0.305 < p < 0.447 
B) 0.311 < p < 0.440 
C) 0.330 < p < 0.421 
D) 0.322 < p < 0.430 
A) 0.305 < p < 0.447 
B) 0.311 < p < 0.440 
C) 0.330 < p < 0.421 
D) 0.322 < p < 0.430

Question 14

Use the confidence level and sample data to find the margin of error E
-College students' annual earnings: $99 \%$ confidence; $n = 71 , \bar { x } = \$ 3660 , \sigma = \$ 879$

Accepted Answer

A)  $\$ 1118$ 
B)  $\$ 243$ 
C)  $\$ 269$ 
D)  $\$ 8$ 
A)  $\$ 1118$ 
B)  $\$ 243$ 
C)  $\$ 269$ 
D)  $\$ 8$

Question 15

Provide an appropriate response.
-What is the best point estimate for the population proportion? Explain why that point estimate is best.

Accepted Answer

The answer of Provide an appropriate response.
-What is the best...

Question 16

Use the margin of error, confidence level, and standard deviation ? to find the minimum sample size required to estimate an unknown population mean µ
-Margin of error: $126, confidence level: 99%, = $534

Accepted Answer

A) 69 
B) 61 
C) 105,268 
D) 120 
A) 69 
B) 61 
C) 105,268 
D) 120

Question 17

Do one of the following, as appropriate: (a) Find the critical value$ z_{?/2}$, (b) find the critical value $ t_{?/2}$, (c) state that neither the normal nor the t distribution applies
-$98 \% ; \mathrm { n } = 7 ; \sigma = 27 ;$ population appears to be normally distributed.

Accepted Answer

A)  $\mathrm { t } _ { \alpha / 2 } = 1.96$ 
B)  $\mathrm { z } _ { \alpha / 2 } = 2.33$ 
C)  $\mathrm { t } _ { \alpha / 2 } = 2.575$ 
D)  $\mathrm { z } _ { \alpha / 2 } = 2.05$ 
A)  $\mathrm { t } _ { \alpha / 2 } = 1.96$ 
B)  $\mathrm { z } _ { \alpha / 2 } = 2.33$ 
C)  $\mathrm { t } _ { \alpha / 2 } = 2.575$ 
D)  $\mathrm { z } _ { \alpha / 2 } = 2.05$

Question 18

Find the margin of error.
-$$\mathrm { n } = 12 , \overline { \mathrm { x } } = 19.1 , \mathrm {~s} = 5.0,99 \text { percent }$$

Accepted Answer

A)  $14.63 < \mu < 23.57$ 
B)  $14.62 < \mu < 23.58$ 
C)  $14.53 < \mu < 23.67$ 
D)  $15.18 < \mu < 23.02$ 
A)  $14.63 < \mu < 23.57$ 
B)  $14.62 < \mu < 23.58$ 
C)  $14.53 < \mu < 23.67$ 
D)  $15.18 < \mu < 23.02$

Question 19

The following confidence interval is obtained for a population proportion, p: (0.298, 0.338) Use these confidence interval limits to find the point estimate, $\hat{\mathrm { p }}$.

Accepted Answer

A)  $0.318$ 
B)  $0.321$ 
C)  $0.298$ 
D)  $0.338$ 
A)  $0.318$ 
B)  $0.321$ 
C)  $0.298$ 
D)  $0.338$

Find the margin of error for the 95% confidence interval used to estimate the population proportion -n = 169, x = 107

Provide an appropriate response. - $\text { Why is } s ^ { 2 } \text { the best point estimate of } \sigma ^ { 2 } \text { ? }$

Find the margin of error - $90 \% ; n = 9 ; \sigma = 4.2 ;$ population appears to be very skewed.

Solve the problem. -Of 243 employees selected randomly from one company, 14.81% of them commute by carpooling. Construct a 90% confidence interval for the true percentage of all employees of the company who carpool.

The following confidence interval is obtained for a population proportion, p: 0.855 < p < 0.897 Use these confidence interval limits to find the margin of error, E.

Use the confidence level and sample data to find the margin of error E -Replacement times for washing machines: $90 \%$ confidence; $\mathrm { n } = 36 , \bar{\mathrm { x} } = 10.0$ years, $\sigma = 2.1$ years

Solve the problem. -Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all such items that are defective.

Solve the problem. -When 306 college students are randomly selected and surveyed, it is found that 115 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.

Use the confidence level and sample data to find the margin of error E -College students' annual earnings: $99 \%$ confidence; $n = 71 , \bar { x } = \$ 3660 , \sigma = \$ 879$

Provide an appropriate response. -What is the best point estimate for the population proportion? Explain why that point estimate is best.

Use the margin of error, confidence level, and standard deviation ? to find the minimum sample size required to estimate an unknown population mean µ -Margin of error: $126, confidence level: 99%, = $534

Do one of the following, as appropriate: (a) Find the critical value $z_{?/2}$ , (b) find the critical value $t_{?/2}$ , (c) state that neither the normal nor the t distribution applies - $98 \% ; \mathrm { n } = 7 ; \sigma = 27 ;$ population appears to be normally distributed.

Find the margin of error. - $\mathrm { n } = 12 , \overline { \mathrm { x } } = 19.1 , \mathrm {~s} = 5.0,99 \text { percent }$

The following confidence interval is obtained for a population proportion, p: (0.298, 0.338) Use these confidence interval limits to find the point estimate, $\hat{\mathrm { p }}$ .

Introduction to Statistics

Summarizing and Graphing Data

Statistics for Describing, Exploring, and Comparing Data

Probability

Probability Distributions

Normal Probability Distributions

Hypothesis Testing

Inferences From Two Samples

Correlation and Regression

Multinomial Experiments and Contingency Tables

Analysis of Variance

Nonparametric Statistics

Statistical Process Control

Filters

Exam 7: Estimates and Sample Sizes

Find the margin of error for the 95% confidence interval used to estimate the population proportion -n = 169, x = 107

Provide an appropriate response. - Why is s2 the best point estimate of σ2 ? \text { Why is } s ^ { 2 } \text { the best point estimate of } \sigma ^ { 2 } \text { ? } Why is s2 the best point estimate of σ2 ?

Find the margin of error - 90%;n=9;σ=4.2;90 \% ; n = 9 ; \sigma = 4.2 ;90%;n=9;σ=4.2; population appears to be very skewed.

Solve the problem. -Of 243 employees selected randomly from one company, 14.81% of them commute by carpooling. Construct a 90% confidence interval for the true percentage of all employees of the company who carpool.

The following confidence interval is obtained for a population proportion, p: 0.855 < p < 0.897 Use these confidence interval limits to find the margin of error, E.

Use the confidence level and sample data to find the margin of error E -Replacement times for washing machines: 90%90 \%90% confidence; n=36,xˉ=10.0\mathrm { n } = 36 , \bar{\mathrm { x} } = 10.0n=36,xˉ=10.0 years, σ=2.1\sigma = 2.1σ=2.1 years

Solve the problem. -Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all such items that are defective.

Solve the problem. -When 306 college students are randomly selected and surveyed, it is found that 115 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.

Use the confidence level and sample data to find the margin of error E -College students' annual earnings: 99%99 \%99% confidence; n=71,xˉ=$3660,σ=$879n = 71 , \bar { x } = \$ 3660 , \sigma = \$ 879n=71,xˉ=$3660,σ=$879

Provide an appropriate response. -What is the best point estimate for the population proportion? Explain why that point estimate is best.

Use the margin of error, confidence level, and standard deviation ? to find the minimum sample size required to estimate an unknown population mean µ -Margin of error: $126, confidence level: 99%, = $534

Find the margin of error. - n=12,x‾=19.1, s=5.0,99 percent \mathrm { n } = 12 , \overline { \mathrm { x } } = 19.1 , \mathrm {~s} = 5.0,99 \text { percent }n=12,x=19.1, s=5.0,99 percent

The following confidence interval is obtained for a population proportion, p: (0.298, 0.338) Use these confidence interval limits to find the point estimate, p^\hat{\mathrm { p }}p^​ .

Introduction to Statistics

Summarizing and Graphing Data

Statistics for Describing, Exploring, and Comparing Data

Probability

Probability Distributions

Normal Probability Distributions

Hypothesis Testing

Inferences From Two Samples

Correlation and Regression

Multinomial Experiments and Contingency Tables

Analysis of Variance

Nonparametric Statistics

Statistical Process Control

Filters

Provide an appropriate response. - $\text { Why is } s ^ { 2 } \text { the best point estimate of } \sigma ^ { 2 } \text { ? }$

Find the margin of error - $90 \% ; n = 9 ; \sigma = 4.2 ;$ population appears to be very skewed.

Use the confidence level and sample data to find the margin of error E -Replacement times for washing machines: $90 \%$ confidence; $\mathrm { n } = 36 , \bar{\mathrm { x} } = 10.0$ years, $\sigma = 2.1$ years

Use the confidence level and sample data to find the margin of error E -College students' annual earnings: $99 \%$ confidence; $n = 71 , \bar { x } = \$ 3660 , \sigma = \$ 879$

Find the margin of error. - $\mathrm { n } = 12 , \overline { \mathrm { x } } = 19.1 , \mathrm {~s} = 5.0,99 \text { percent }$

The following confidence interval is obtained for a population proportion, p: (0.298, 0.338) Use these confidence interval limits to find the point estimate, $\hat{\mathrm { p }}$ .